Change the order of integration for triple integral.

[Baron]

How do I write the below integral in the order of dzdxdy?

\(\displaystyle \int_{-1}^{1}\;\ \int_{0}^{1-y^2}\;\ \int_{0}^{2-y-z}dxdzdy\)

I'm having difficulties because the limits of x are 0 ≤ x ≤ 2 - y -z . There are two variables for x so I can't use Fubini's Theorem.

For the order of dzdxdy, I think the limits of integration of y would stay the same as -1 ≤ y ≤ 1

I'm pretty sure the limits of integration for z would be 0 ≤ z ≤ 2 - y - x, the upper limit of integration is found by isolating z from the equation x = 2 - y - z .

Is my work correct so far? How do I find the limits of integration for x? I think the bottom limit of integration is 0 from the x = 0 part of the original integral. But what is the upper limit?

 

[Deleted member 4993]

How do I write the below integral in the order of dzdxdy?

\(\displaystyle \int_{-1}^{1}\;\ \int_{0}^{1-y^2}\;\ \int_{0}^{2-y-z}dxdzdy\)

I'm having difficulties because the limits of x are 0 ≤ x ≤ 2 - y -z . There are two variables for x so I can't use Fubini's Theorem.

For the order of dzdxdy, I think the limits of integration of y would stay the same as -1 ≤ y ≤ 1

I'm pretty sure the limits of integration for z would be 0 ≤ z ≤ 2 - y - x, the upper limit of integration is found by isolating z from the equation x = 2 - y - z .

Is my work correct so far? How do I find the limits of integration for x? I think the bottom limit of integration is 0 from the x = 0 part of the original integral. But what is the upper limit?

Did you try to visualize the shape of the volume that you are interested in?

What are the shapes of those limits? x+y+z = 2 is a slanted plane. z = 1-y² is a paraboloid..

Once you have that clarified in your mind - limit changing would be easy.

 

[HallsofIvy]

If you had "visualized" or, better, drawn, the figure, you would have noticed a fundamental problem. x goes from 0 to 2- y- z. In the y= 0 plane the boundary is x= 2- z and when x= 0, z= 2. But y goes from 0 to \(\displaystyle 1- z^2\) so that when y= 0, \(\displaystyle z= \pm 1\). The boundaries do not connect and this integral is NOT over a complete closed figure. Check the original problem and see it it was not either "x from 0 to 1- y- z" or "y from 0 to \(\displaystyle 4- x^2\)".

 

[Baron]

Did you try to visualize the shape of the volume that you are interested in?

What are the shapes of those limits? x+y+z = 2 is a slanted plane. z = 1-y² is a paraboloid..

Once you have that clarified in your mind - limit changing would be easy.

I have difficulty visualizing the region especially since the limits of x contain two variables.
x + y + z = 2 intersects the x, y, z axis at x = 2, y = 2, z = 2 respectively.
z = 1 - y^2 is a parabola in the yz axis where y is between -1 and 1
and x = 0 and z = 0, meaning the region is found in the 1st quadrant and the quadrant where y is negative and x and z are both positive.

But I have no idea what the boundaries are mainly because of what HallsofIvy said.

If you had "visualized" or, better, drawn, the figure, you would have noticed a fundamental problem. x goes from 0 to 2- y- z. In the y= 0 plane the boundary is x= 2- z and when x= 0, z= 2. But y goes from 0 to \(\displaystyle 1- z^2 \) so that when y= 0, \(\displaystyle z= \pm 1\). The boundaries do not connect and this integral is NOT over a complete closed figure. Check the original problem and see it it was not either "x from 0 to 1- y- z" or "y from 0 to \(\displaystyle 4- x^2\)".

I've attached a copy of the original problem. But how does y go from 0 to 1 - z^2? I thought z = 1 - y^2 so y = ±sqrt(1-z). From the diagram I drew, I think y goes from -sqrt(1-z) to either sqrt(1-z) or 2 - x - z depending on what the boundaries are.